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Let $p_1,\dots , p_{n+1}$ be distinct primes, let $\alpha_1, \dots , \alpha_n$ be integers, and let $a,b$ be integers. Suppose we had the equation:

$$b^2p_{n+1} = a^2p_1^{\alpha_1}\dots p_n^{\alpha_n}$$ my book states that this is not possible, because the highest power of $p_{n+1}$ dividing the left hand side is odd, whereas the highest power of $p_{n+1}$ dividing the right hand side is even. Could someone explain this argument of "dividing the highest power of $p_{n+1}$

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  • $\begingroup$ Suppose the highest power of $p_{n+1}$ dividing $b$ is $p_{n+1}^r$, and the highest power of $p_{n+1}$ dividing $a$ is $p_{n+1}^s$. Then the highest power dividing $b^2p_{n+1}$ is $p_{n+1}^{2r+1}$, which is an odd power, whereas the highest power dividing $a^2p_1^{\alpha_1}\dots p_n^{\alpha_n}$ is $p_{n+1}^{2s}$, which is an even power. $\endgroup$ – almagest May 12 '16 at 14:05
  • $\begingroup$ Definitions are your friends. A prime is a ring element $p$ such that $p \mid rs$ implies either $p\mid r$ or $p\mid s$. $\endgroup$ – hardmath May 12 '16 at 14:35
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The fundamental theorem of arithmetic says that every positive integer has a unique factorization into a product of primes (up to reordering). In the prime factorization of $b^2p_{n+1}$, the prime $p_{n+1}$ has some exponent, say $k$. That is, $p_{n+1}^k$ divides $b^2p_{n+1}$ but no higher power of $p_{n+1}$ does. As $b^2p_{n+1}=a^2p_1^{\alpha_1}\cdots p_n^{\alpha_n}$, we know $p_{n+1}^k$ is also the highest power of $p_{n+1}$ that divides the right-hand side of the equation.

Now the claim is that $k$ must be both even and odd, leading to a contradiction. Consider the left-hand side of the equation. Let $p_{n+1}^\ell$ be the highest power of $p_{n+1}$ dividing $b$. Then $p_{n+1}^{2\ell}$ is the highest power of $p_{n+1}$ dividing $b^2$, and thus $p_{n+1}^{2\ell+1}$ is the highest power of $p_{n+1}$ dividing $b^2p_{n+1}$. That is, $k=2\ell+1$, so $k$ is odd.

Next consider the right-hand side. We know $p_{n+1}$ does not divide any of the primes $p_1,\dots,p_n$. Let $p_{n+1}^m$ be the highest power of $p_{n+1}$ dividing $a$. Then $p_{n+1}^{2m}$ is the highest power of $p_{n+1}$ dividing $a^2$, and thus also the highest power of $p_{n+1}$ dividing $a^2p_1^{\alpha_1}\cdots p_n^{\alpha_n}$. That is, $k=2m$, so $k$ is even, a contradiction.

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